a.) To determine the angle of projection, we can use the formula:
θ = tan⁻¹(v₀y / v₀x)
where θ is the angle of projection, v₀y is the initial vertical velocity component, and v₀x is the initial horizontal velocity component.
Given that the initial upward component is 30m/s and the horizontal component is 20m/s, we have:
θ = tan⁻¹(30 / 20) ≈ 56.31°
Therefore, the angle of projection is approximately 56.31°.
b.) To find the velocity magnitude after 5 seconds, we need to find the components of the velocity at that time. The vertical component will be affected by gravity, so we can use:
v_y = v₀y + gt
where v_y is the vertical velocity component at time t, g is the acceleration due to gravity (9.81 m/s²), and t is the time in seconds.
We have:
v_y = 30 - 9.81(5) = -19.05 m/s
The negative sign indicates that the bullet is moving downwards at that time. The horizontal component, however, remains constant at 20 m/s.
The velocity magnitude can be found using the Pythagorean theorem:
v = √(v_x² + v_y²)
where v_x is the horizontal velocity component and v_y is the vertical velocity component.
We have:
v = √(20² + (-19.05)²) ≈ 27.14 m/s
Therefore, the velocity magnitude after 5 seconds is approximately 27.14 m/s.
c.) To find the time required to return to the original level, we need to consider the vertical motion of the bullet. At its highest point, the vertical velocity component will be zero. We can use:
v_y = v₀y + gt
to find the time at which the bullet reaches its highest point.
We have:
0 = 30 - 9.81t
t = 30 / 9.81 ≈ 3.06 s
Therefore, it takes approximately 3.06 seconds for the bullet to reach its highest point. It will take another 3.06 seconds to return to the original level, for a total time of 6.12 seconds.
d.) To find the range, we need to consider the horizontal motion of the bullet. The range is the distance travelled in the horizontal direction before the bullet hits the ground. We can use:
x = v₀x t
where x is the range, v₀x is the initial horizontal velocity component, and t is the time of flight.
The time of flight can be found using:
t = 2v₀y / g
where v₀y is the initial vertical velocity component and g is the acceleration due to gravity.
We have:
t = 2(30) / 9.81 ≈ 6.12 s
Therefore, the time of flight is approximately 6.12 seconds, which we already found in part c.
Now we can find the range:
x = (20)(6.12) = 122.4 m
Therefore, the range of the bullet is approximately 122.4 meters.
PH